MI PU2 Promo 9758/2019/02/Q1 Town A, Town B and Town C are located in Wakandi Country. The distance between Town A and Town B is 60 km and the distance between Town B and Town C is 20 km. A railroad connects Town A to Town B (see diagram). A manufacturer plans to deliver a certain number of containers of its goods daily from Town A to Town C. To support this plan, the Wakandi government decides to build Interchange Station S and a road connecting this station to Town C (see diagram). Once the road is built, the goods manufacturer can deliver its containers from Town A to Town C by a combination of rail and road via Interchange Station S. The cost to deliver the containers daily by rail is $200 per km and the cost to deliver the containers daily by road is$300 per km.

1. Find the distance SB and use Pythagoras' theorem to express the road distance SC in terms of $x$.

Since the distance from A to B is 60 km and $AS = x$ km,
$SB = 60 - x$

Using Pythagoras' theorem on the right-angled triangle $SBC$,
$SC^2 = SB^2 + BC^2$
$SC = \sqrt{(60 - x)^2 + 20^2}$
$SC = \sqrt{3600 - 120x + x^2 + 400}$
$SC = \sqrt{x^2 - 120x + 4000}$

2. Calculate the total cost by combining the cost of rail and road delivery.

Cost for rail delivery from A to S $= 200 \times AS = 200x$
Cost for road delivery from S to C $= 300 \times SC = 300\sqrt{x^2 - 120x + 4000}$

Total daily delivery cost, $T = 200x + 300\sqrt{x^2 - 120x + 4000}$ (Shown)

1. Differentiate $T$ with respect to $x$ using the chain rule.

$T = 200x + 300(x^2 - 120x + 4000)^{\frac{1}{2}}$
$\frac{dT}{dx} = 200 + 300\left(\frac{1}{2}\right)(x^2 - 120x + 4000)^{-\frac{1}{2}}(2x - 120)$
$\frac{dT}{dx} = 200 + \frac{150(2x - 120)}{\sqrt{x^2 - 120x + 4000}}$
$\frac{dT}{dx} = 200 + \frac{300(x - 60)}{\sqrt{x^2 - 120x + 4000}}$

2. Equate the derivative to zero and rearrange to isolate the square root term.

For a stationary value, set $\frac{dT}{dx} = 0$:
$200 + \frac{300(x - 60)}{\sqrt{x^2 - 120x + 4000}} = 0$
$\frac{300(x - 60)}{\sqrt{x^2 - 120x + 4000}} = -200$
$\frac{3(x - 60)}{\sqrt{x^2 - 120x + 4000}} = -2$
$3(x - 60) = -2\sqrt{x^2 - 120x + 4000}$
Note: For this equation to hold, the left-hand side must be negative (since the principal square root is always positive), which implies $x < 60$.

3. Square both sides, solve the resulting quadratic equation, and reject the extraneous root.

$9(x - 60)^2 = 4(x^2 - 120x + 4000)$
$9(x^2 - 120x + 3600) = 4x^2 - 480x + 16000$
$9x^2 - 1080x + 32400 = 4x^2 - 480x + 16000$
$5x^2 - 600x + 16400 = 0$
$x^2 - 120x + 3280 = 0$
Using the quadratic formula:
$x = \frac{-(-120) \pm \sqrt{(-120)^2 - 4(1)(3280)}}{2(1)}$
$x = \frac{120 \pm \sqrt{14400 - 13120}}{2}$
$x = \frac{120 \pm \sqrt{1280}}{2} = 60 \pm 8\sqrt{5}$
Since we established earlier that $x < 60$, we reject $x = 60 + 8\sqrt{5}$.
$x = 60 - 8\sqrt{5} \approx 42.111 \approx 42.11 \text{ (to 2 d.p.)}$

4. Use the second derivative test to verify that this value of $x$ yields a minimum cost.

$\frac{d^2T}{dx^2} = \frac{d}{dx} \left( 200 + 300\frac{x - 60}{\sqrt{x^2 - 120x + 4000}} \right)$
Using the quotient rule:
$\frac{d^2T}{dx^2} = 300 \left[ \frac{\sqrt{x^2 - 120x + 4000}(1) - (x - 60)\left( \frac{1}{2}(x^2 - 120x + 4000)^{-\frac{1}{2}}(2x - 120) \right)}{x^2 - 120x + 4000} \right]$
$= 300 \left[ \frac{\sqrt{x^2 - 120x + 4000} - \frac{(x - 60)^2}{\sqrt{x^2 - 120x + 4000}}}{x^2 - 120x + 4000} \right]$
Multiply the numerator and denominator by $\sqrt{x^2 - 120x + 4000}$:
$= 300 \left[ \frac{(x^2 - 120x + 4000) - (x^2 - 120x + 3600)}{(x^2 - 120x + 4000)^{\frac{3}{2}}} \right]$
$= \frac{120000}{(x^2 - 120x + 4000)^{\frac{3}{2}}}$
Since $(x^2 - 120x + 4000) = (x - 60)^2 + 400 > 0$, we have $\frac{d^2T}{dx^2} > 0$ for all real $x$.
Thus, $T$ is a minimum at $x \approx 42.11$.